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109.67 Derived base change

Let $R \to R'$ be a ring map. In More on Algebra, Section 15.60 we construct a derived base change functor $- \otimes _ R^\mathbf {L} R' : D(R) \to D(R')$. Next, let $R \to A$ be a second ring map. Picture

\[ \xymatrix{ A \ar[r] & A \otimes _ R R' \ar@{=}[r] & A' \\ R \ar[u] \ar[r] & R' \ar[u] \ar[ur] } \]

Given an $A$-module $M$ the tensor product $M \otimes _ R R'$ is a $A \otimes _ R R'$-module, i.e., an $A'$-module. For the ring map $A \to A'$ there is a derived functor

\[ - \otimes _ A^\mathbf {L} A' : D(A) \longrightarrow D(A') \]

but this functor does not agree with $- \otimes _ R^\mathbf {L} R'$ in general. More precisely, for $K \in D(A)$ the canonical map

\[ K \otimes _ R^{\mathbf{L}} R' \longrightarrow K \otimes _ A^{\mathbf{L}} A' \]

in $D(R')$ constructed in More on Algebra, Equation ( isn't an isomorphism in general. Thus one may wonder if there exists a “derived base change functor” $T : D(A) \to D(A')$, i.e., a functor such that $T(K)$ maps to $K \otimes _ R^\mathbf {L} R'$ in $D(R')$. In this section we show it does not exist in general.

Let $k$ be a field. Set $R = k[x, y]$. Set $R' = R/(xy)$ and $A = R/(x^2)$. The object $A \otimes _ R^\mathbf {L} R'$ in $D(R')$ is represented by

\[ x^2 : R' \longrightarrow R' \]

and we have $H^0(A \otimes _ R^\mathbf {L} R') = A \otimes _ R R'$. We claim that there does not exist an object $E$ of $D(A \otimes _ R R')$ mapping to $A \otimes _ R^\mathbf {L} R'$ in $D(R')$. Namely, for such an $E$ the module $H^0(E)$ would be free, hence $E$ would decompose as $H^0(E)[0] \oplus H^{-1}(E)[1]$. But it is easy to see that $A \otimes _ R^\mathbf {L} R'$ is not isomorphic to the sum of its cohomology groups in $D(R')$.

Lemma 109.67.1. Let $R \to R'$ and $R \to A$ be ring maps. In general there does not exist a functor $T : D(A) \to D(A \otimes _ R R')$ of triangulated categories such that an $A$-module $M$ gives an object $T(M)$ of $D(A \otimes _ R R')$ which maps to $M \otimes _ R^\mathbf {L} R'$ under the map $D(A \otimes _ R R') \to D(R')$.

Proof. See discussion above. $\square$

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